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We say a monoidal category $\mathcal V=(\mathcal V_0,\otimes,I,a,l,r)$ is closed if the endofunctor $-⊗Y$ has a right adjoint $[Y,-]$, called the exponential, for every $Y$. The object $[Y,Z]$ for objects $Y,Z\in\mathcal V_0$ is also called internal hom of $Y$ and $Z$. This suggests one can make it a bifunctor $\mathcal V_0^{\text{op}}\times\mathcal V_0\to\mathcal V_0$. Indeed, given any $f:Y'\to Y$, one may define $[f,Z]:[Y,Z]\to[Y',Z]$ as the unique arrow satisfying $$\varepsilon^{Y'}_Z([f,Z]⊗1)=ε^Y_Z(1⊗f)$$ where $e^Y:[Y,Z]⊗Y\Rightarrow Z$ denotes the counit of the adjunction given by $Y$. It is not difficult to show that this defines a contravariant functor $[-,Z]$ for fixed $Z$.

However, in order for $[-,-]$ to be a bifunctor, it needs to satisfy the compatibility condition $$[Y',g][f,Z] = [f,Z'][Y,g] : [Y,Z]\to[Y',Z']\qquad (1)$$ Now one can compute $$\begin{align} ε([Y',g]⊗1)([f,Z]⊗1) & = gε([f,Z]⊗1) \\ & = gε(1⊗f) \\ & = ε([Y,g]⊗1)(1⊗f) \\ & = ε(1⊗f)([Y,g]⊗1) \\ & = ε([f,Z']⊗1)([Y,g]⊗1) \end{align} $$ as arrows $[Y,Z]⊗ Y' \to Z'$ Unless I've made some mistake, this verifies that their adjuncts satisfy (1).

However, I found it nowhere mentioned that $[-,-]$ is really a bifunctor, neither in literature nor in the nLab. Could someone confirm that my reasoning is correct and/or give a reference for this fact? Thanks.

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    $\begingroup$ It is well known and considered obvious. If you really want a citeable justification, look at Theorem 3 in [CWM, Ch. IV, §7]. $\endgroup$ – Zhen Lin May 14 '15 at 18:44
  • $\begingroup$ And for a generalization to n-variables, see Theorem 2.2 here. $\endgroup$ – tcamps May 14 '15 at 19:38
  • $\begingroup$ @ZhenLin: I wouldn't call it obvious. It requires some verification, and while I was trying to figure out the necessary diagram chasing, I wasn't sure if it's true in the first place. I would have expected it to be mentioned somewhere. But thanks for the reference, I'll look into it in order to see how this is a special case. $\endgroup$ – Stefan Hamcke May 14 '15 at 20:56
  • $\begingroup$ Ah, I see Mac Lane mentiones this in [CWM, Ch. IV, §6] for Cartesian closed categories. $\endgroup$ – Stefan Hamcke May 15 '15 at 13:40

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